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I have a problem with understanding the following exercise.

Consider a hyperbolic line l in the hyperboloid model given by l={x $\in P(R^{2,1})$|$\langle x,n\rangle =0, \langle x,x\rangle =-1 $ and $x_3 \gt 3$} where n$\in P(R^{2,1})$ with $\langle n,n\rangle =1$ is the unit normal of the line. In the Poincaré half-plane model $H^2$ the hyperbolic line l is represented by an arc of the circle $S_l$ orthogonal to the boundary of the halfplane. Show that the (hyperbolic) reflection in l is given by the inversion in the circle $S_l$.

I do know that the reflection with respect to l is given by $x \to x'=x-2*\langle n,x \rangle'n$

while the inversion in the circle $S_l$ is given by $z \to z'=c+\frac{r^2}{|z-c|^2}(z-c)$ where c denotes the center and r denotes the radius of $S_l$.

In an informal sense I have to show that both maps have the same effect on points in their respective models but I do not understand what I have to show to give a proof that is formally correct.

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  • $\begingroup$ I think that should be $x_3>0$. $\endgroup$ – mr_e_man Sep 8 '18 at 20:10
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The key property is that if you reflect a point and then switch model, or switch model and then reflect, the result is the same.

Formally speaking I'd start by defining some mappings. Something like this:

Let $\phi:\mathbb P(\mathbb R^{2,1})\to\mathbb H^2$ be the map which takes points from the hyperboloid model to the half-plane model. Let $\phi'$ be the corresponding map for lines. Let $\rho_l:\mathbb P(\mathbb R^{2,1})\to\mathbb P(\mathbb R^{2,1})$ be a reflection in $l$ in the Hyperbolid model and let $\rho'_{l'}:\mathbb H^2\to\mathbb H^2$ be an inversion in $l'$ the half-plane model.

Using these symbols, the property I stated at the beginning can be formalized like this:

$$\phi(\rho_l(p)) = \rho'_{\phi'(l)}(\phi(p))$$

Depending on how much you trust your formulas, you might want to check additional properties. For example you could check that $\phi$ and $\phi'$, i.e. the mapping of points and that of lines, preserve incidence: if you map a line and a point on it, the image point will lie on the image line.

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